If the lines frac{x-1}{2}=frac{y+1}{3}=frac{z | vedclass

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(D) Let the given lines be:$L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda \implies x = 2\lambda+1, y = 3\lambda-1, z = 4\lambda+1$$L_2: \fr

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(D) Let the given lines be:$L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda \implies x = 2\lambda+1, y = 3\lambda-1, z = 4\lambda+1$$L_2: \frac{x-2}{1}=\frac{y+m}{2}=\frac{z-2}{1} = \mu \implies x = \mu+2, y = 2\mu-m, z = \mu+2$Since the lines intersect,there exists a point $(x, y, z)$ common to both lines.Equating the $x$ and $z$ coordinates:$2\lambda+1 = \mu+2 \implies 2\lambda - \mu = 1$  $(1)$$4\lambda+1 = \mu+2 \implies 4\lambda - \mu = 1$  $(2)$Subtracting $(1)$ from $(2)$,we get $2\lambda = 0 \implies \lambda = 0$.Substituting $\lambda = 0$ in $(1)$,we get $-\mu = 1 \implies \mu = -1$.Now,equate the $y$ coordinates:$3\lambda - 1 = 2\mu - m$Substituting $\lambda = 0$ and $\mu = -1$:$3(0) - 1 = 2(-1) - m$$-1 = -2 - m$$m = -2 + 1 = -1$.

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-2}{1}=\frac{y+m}{2}=\frac{z-2}{1}$ intersect each other,then the value of $m$ is

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